Light form a dicharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photoelectrons emitted from sodium is 0.73 eV. The work function for sodium is 1.82 eV. Find (a) the energy of the photons causing the photoelectrons emission.
(b) the quantum numbers of the two levels involved in the emission of these photons.
(c ) the change in the angular momentum of the electron in the hydrogen atom, in the above transition, and
(d) the recoil speed of the emitting atom assuming it to be at rest before the transition. (lonization potential of hydrogen is 13.6 eV.)

Explanation

(i) According to Einstein 's photo -electric equation, the maximum kinetic energy `E_(k)` of the emitted electrons is given by `(E)_(K)^(max)=hv-`W`,wgere hv is the energy of photons causing the photo -electric emission and W is the work function of the emitting surface.Given that, `(E)_(k)^(max)=0.73 eV and W= 1.82 eV `, `therefore` `hv= (E)_(k)^(max)+W =0.73 eV +1.82 eV =2.55eV` (ii) These photons (whose energy is 2.55 eV) are by hydrogen atoms.As `(I.E)_(H)` =1.36eV`, hence `(E)_(1)^(H)=-(I.E)_(H)=-13.6eV` The energy of higher levels is given by `E_(n)^(H)=(E_(1)^(H))/(n^(2)`.Hence ,`E_(2)^(H)=-(13.6)/(4)=-3.4 eV,E_(3)^(H)= -(13.6)/(9)=-1.5eV` and `E_(4)^(H)= -(13.6)/(16)=-0.85eV`.The energy of the emitted photon is 2.55 eV.Now `E_(4)^(H)-E_(2)^(H)=-0.85eV -(3.4eV)=2.55eV`.Thus the quantum numbers of two levls involved in the emission of photon of energy 2.55eV are 4 and 2.(iii) The elctron transition causing the emission of photon of eneergy 2.55eV is from n=4 level to n=2 level.Now, according tp Bohr's `2^(nd)` postulate,the angular momentum of electron in the hydrogen atom is `((nh)/(2pi))`.Thus, the change in angular momentum in the above transition is `Delta L =(4h)/2pi)-(2h)/(2pi)=(h)/(pi)` (iv) The momentum of the photon emitted from the hydrogen atom `p_(ph)=(hv)/(c)=(2.55xx(1.6xx10^(-18))J)/(3xx10^(8)m//s)=1.36xx10^(-27)kg .m//s` According to the law of conservation of momentum , the recoil momentum of a hydrogen atom will be equal and opposite to the momentum of the emitted photon.`(vecp_(ph)+vecpA.or `vec p_A=-vecp_(ph))`Hence the recoil speed of the atom is `v=|Momentum|/|mass|=|vecp_A|/(m_A)=|vecp_(ph)|/|m_A|`=`((1.36xx10^(-27))kg-m//s)/((1.67xx10^(-27))kg)=0.814m//s`